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प्रश्न

\[\int\limits_0^{\pi/2} \cos^2 x\ dx\]

उत्तर

\[Let I = \int_0^\frac{\pi}{2} \cos^2 x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^2 x\ d\ x\]
\[ \Rightarrow I = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 + \cos 2x \right) dx \left[ \because \cos 2x = 2 \cos^2 x - 1 \right]\]
\[ \Rightarrow I = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi}{4} + 0 - 0\]
\[ \Rightarrow I = \frac{\pi}{4}\]

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Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.1 | Q 17 | पृष्ठ १६

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